Problem: The equation of a circle $C$ is $x^2+y^2-10x-12y+12 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Solution: To find the equation in standard form, complete the square. $(x^2-10x) + (y^2-12y) = -12$ $(x^2-10x+25) + (y^2-12y+36) = -12 + 25 + 36$ $(x-5)^{2} + (y-6)^{2} = 49 = 7^2$ Thus, $(h, k) = (5, 6)$ and $r = 7$.